Chemistry, asked by vinusha, 1 year ago

at 353 kelvin the ph of 0.001 M KOH will be

Answers

Answered by soumyamehta26

At 363, Kw = 3.80X10^-13

So, Kw = [H+][OH-] = 3.80X10^-13

In this solution, [OH-] = 0.001, so:

[H+] = 3.8X10^-13 / 0.001 = 3.80X10^-10

pH = -log [H+] = 9.4

vinusha: is it necessary to know the Kw prerequisite value prerequisite for this problem?

vinusha: is it necessary to know the Kw value prerequisite for this problem?

soumyamehta26: Like any other equilibrium constant, the value of Kw varies with temperature. Its value is usually taken to be 1.00 x 10-14 mol2 dm-6 at room temperature. In fact, this is its value at a bit less than 25°C. But for other temperatures, we have to know the value beforehand. :)

vinusha: thank you

soumyamehta26: My pleasure :)

Answered by Light1Yagami

Answer:

Explanation:

Here's the answer for your question

Ans is 9 and the explanation is below

Attachments:

Previous Question

Next Question